Integration of $ \int_{0}^{\frac{\pi}{2}} x \log(1-\cos x) \,dx $

Question

Question: What is the closed form of this following integral? $ \int_{0}^{\frac{\pi}{2}} x \log(1-\cos x) \,dx.$

We have $ a_{k}:=\int_{0}^{\pi/2}{x \cos^k x dx},$ then posit from the Beppo-Levi theorem, we have $ \int_{0}^{\pi/2}{x\ln\left(1-2\cos x \right)}dx=-\sum_{k=1}^{\infty}{\frac{a_{k}}{k}}.$

$ a_{k}=\int_{0}^{\pi/2}{x\cos^{k-1}x \left(\sin x \right)x}=-1/k -\left(k-1 \right)a_{k}-\left(k-1 \right)a_{k-2}.$

From here we have the reductionist relation.With $\displaystyle ka_{k}-\left(k-1 \right)a_{k-2}=\frac{-1}{k}$ a simple induction we have that $ a_{k}=b_{k}\pi^2 +c_{k}\pi+d_{k},a_{k},b_{k},c_{k} \in \mathbb{Q}.$ From here we easily find that $2kb_{2k}=(2k-1)b_{2k-2},b_{2k+1}=0,b_{2}=1/8$.Therefore $ b_{2k}=\frac{1}{8}\cdot \frac{\binom{2k}{k}}{4^{k}}$

Consequently $ \sum_{k=1}^{\infty}{\frac{b_{k}}{k}}=\frac{1}{16}\sum_{k=1}^{\infty}{\frac{1}{k\cdot 4^{k}}\binom{2k}{k}}=\frac{1}{16}\int_{0}^{1/4}{\left(\frac{1}{\sqrt{1-4x}}-1 \right)\frac{1}{x}}dx= \frac{1}{16}\int_{0}^{1}{\frac{1}{\sqrt{x}\left(\sqrt{x}+1 \right)}}dx=...=\frac{\ln 2}{8}.$

For the sequence $c_{k}$ it easily follows that , so we have that $(2k+1)a_{2k+1}=2ka_{2k-1},a_{1}=1/2\Rightarrow a_{2k+1}=\frac{4^{k}\left(k!)^2 \right)}{2\left(2k \right)!(2k+1)}$= $ \sum_{k=1}^{\infty}{\frac{c_{k}}{k}}=\sum_{n=0}^{\infty}{\frac{4^{n}\left(n! \right)^2}{2\cdot (2n)!(2n+1)^2}}.$

Answer 1

Using $$ \Re\log(1-e^{ix})=\ln|1-e^{ix}|=\frac12(\log2+\log(1-\cos x)) $$ one has $$\begin{eqnarray} I&=&\int_{0}^{\frac{\pi}{2}} x \log(1-\cos x) \,dx\\ &=&-\frac18\pi^2\log2+2\Re\int_{0}^{\frac{\pi}{2}} x \log(1-e^{ix}) \,dx\\ &=&-\frac18\pi^2\log2+2\Re i\int_{0}^{\frac{\pi}{2}} x d\text{Li}_2\left(e^{i x}\right)\\ &=&-\frac18\pi^2\log2+2\Re \bigg[i x \text{Li}_2\left(e^{i x}\right)-\text{Li}_3\left(e^{i x}\right)\bigg]\bigg|_{0}^{\frac{\pi}{2}}\\ &=&-\frac18\pi^2\log2+2\Re \bigg[\frac\pi2 i\text{Li}_2(i)-\text{Li}_3(i)+\zeta(3)\bigg]. \end{eqnarray}$$ Update: (1) It is easy to see $$ i\text{Li}_2(i)=-C-i\frac{\pi^2}{48}. $$ (2). From https://en.wikipedia.org/wiki/Polylogarithm, one has $$ \text{Li}_3(i)=-2^{-3}\eta(3)+i\beta(3)=-\frac18(1-\frac1{2^2})\zeta(3)+i\beta(3)=-\frac3{32}\zeta(3)+i\beta(3). $$ So $$ I=-\frac{\pi^2}8\ln2+\frac{35}{16}\zeta(3)-\pi C $$

Answer 2

\begin{align} &\int_{0}^{\frac{\pi}{2}} x \ln(1-\cos x) \,dx \\ = & \int_{0}^{\frac{\pi}{2}} x \ln(\sin x) \overset{x\to\frac \pi2-x}{dx } + \int_{0}^{\frac{\pi}{2}} x\ln(\tan\frac x2)\,dx \\ = &\ \frac12\int_{0}^{\frac{\pi}{2}} \left[x \ln(\sin x) +(\frac\pi2-x)\ln(\cos x)\right]dx\\ & + \int_{0}^{{\pi}} x\ln(\tan\frac x2) \overset{x\to 2x}{dx } -\int_{\frac\pi2}^{{\pi}} x\ln(\tan\frac x2) \overset{x\to\pi-x}{dx } \\ =& \ \frac52\int_{0}^{\frac{\pi}{2}} x\ln(\tan x)dx + \frac\pi4\int_{0}^{\frac{\pi}{2}} \ln(\cos x) dx + \frac\pi2 \int_0^{\frac\pi2}\ln(\tan \frac x2)dx\\ =& \ \frac{35}{16}\zeta(3) -\frac{\pi^2}8\ln2-\pi G \end{align}

where $ \int_{0}^{\frac{\pi}{2}} \ln(\cos x) dx=-\frac\pi2\ln2$, $ \int_0^{\frac\pi2}\ln(\tan \frac x2)dx =-2G$ and $\int_{0}^{\frac{\pi}{2}} x\ln(\tan x)= \frac78\zeta(3)$.

Answer 3

First, we have

$$\begin{align} \log(1-\cos(x))&=\log\left(2\sin^2(x/2)\right)\\\\ &=\log(2)+2\log(\sin(x/2))\\\\ &=\log(2)+2\left(-\log(2)-\sum_{n=1}^\infty \frac{\cos(nx)}{n}\right)\\\\ &=-\log(2)-2\sum_{n=1}^\infty \frac{\cos(nx)}{n} \end{align}$$

where we used the well-known Fourier Series (See Here for example) of $\log(\sin(x/2))$.

Then, we have

$$\begin{align} \int_0^{\pi/2}x\log(1-\cos(x))\,dx&=-\frac{\pi^2}8 \log(2)-2\sum_{n=1}^\infty \frac1n\int_0^{\pi/2}x\cos(nx)\,dx\\\\ &=-\frac{\pi^2}8 \log(2)+2\sum_{n=1}^\infty \left(\frac1{n^3}-\frac\pi2 \frac{\sin(n\pi/2)}{n^2}-\frac{\cos(n\pi/2)}{n^3}\right)\\\\ &=-\frac{\pi^2}8 \log(2)+2\sum_{n=1}^\infty \frac1{n^3}+\pi \sum_{n=1}^\infty \frac{(-1)^{n}}{(2n-1)^2}+2\sum_{n=1}^\infty \frac{(-1)^{n-1}}{(2n)^3}\\\\ &=-\frac{\pi^2}8 \log(2)+2\zeta(3)-\pi G+\frac3{16}\zeta(3)\\\\ &=-\frac{\pi^2}8 \log(2)+\frac{35}{16}\zeta(3)-\pi G \end{align}$$

where we used $\sum_{n=1}^\infty \frac{1}{n^3}=\zeta(3)$, $\sum_{n=1}^\infty \frac{(-1)^n}{(2n-1)^2}=-G$, and $\sum_{n=1}^\infty \frac{(-1)^{n-1}}{(2n)^3}=\frac{3}{4}\zeta(3)$.